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thanksLet theta = @
Let omega = w
View attachment 57781
SQ is clearly r(sin@)
So SQ = rsin(@/t)*t Where @/t = w
a)PLZ help with this !View attachment 57783
No...Actually the centripetal force acting toward the centre is approximately at 90 degrees of an angle to the velocity in which the object is moving.Ok yeah taking the real case scenario it is clear
But then as you said "a 'centripetal' force is any force that acts towards the centre. since the mud is on an object in circular motion, it will experience a force."
Doesn't that mean the mud will experience a force towards the centre?
Total current = 230/44.6
maxm V = 2V (I = 2/3.4 ; V = 2 * 3.4/3.4)
I dont get it.maxm V = 2V (I = 2/3.4 ; V = 2 * 3.4/3.4)
Minm V = 4.5 x 10^-3 (I = 2/(1500+3.4) ; V = I x 3.4)
Maximum V is when R = 0ohms.I dont get it.
I checked there first.I did not understand.
3.8 x 10^5 will be the distance of the edge of moon to telescope (radius)
I checked there first.I did not understand.
Dont u get it still?Maximum V is when R = 0ohms.
So I when R = oohms is 2/3.4 A
So maximum V = I * R
R = 3.4
So V = 2/3.4 x 3.4 = 2V
Minimum V when R = 150ohms.
so I = 2/(1500 + 3.4) A
So V = 2/1503.4 x 3.4 = 4.5 x 10^-3V
I think I do.thanksDont u get it still?
Exactly which part you are not getting? I will try to simplify that.I think I do.thanks
No let me try once more,later.Exactly which part you are not getting? I will try to simplify that.
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